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# Help? Complex numbers

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Help? Complex numbers

Oct 3, 2017

#2
+97500
+2

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

$$x=\frac{3\pm\sqrt{9-8j}}{2}$$

The roots will be complex when  the discriminant is negative.

That is

$$9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125$$

So the solutions (for x) of  x^2-3x+2j=0    will be complex if   j>1.125

Oct 3, 2017

#1
+1

I replaced your "j" with "i", which is a more commonly used "complex number"

Solve for x:
2 i - 3 x + x^2 = 0

Subtract 2 i from both sides:
x^2 - 3 x = -2 i

x^2 - 3 x + 9/4 = 9/4 - 2 i

Write the left hand side as a square:
(x - 3/2)^2 = 9/4 - 2 i

Take the square root of both sides:
x - 3/2 = sqrt(9/4 - 2 i) or x - 3/2 = -sqrt(9/4 - 2 i)

x = sqrt(9/4 - 2 i) + 3/2 or x - 3/2 = -sqrt(9/4 - 2 i)

x = 3/2 + 1/2 sqrt(9 - 8 i)      or      x = 3/2 - sqrt(9/4 - 2 i)

Oct 3, 2017
#2
+97500
+2

I think you have misunderstood the question guest ://

j is not a complex number.

x^2-3x+2j=0

Solutions for x

$$x=\frac{3\pm\sqrt{9-8j}}{2}$$

The roots will be complex when  the discriminant is negative.

That is

$$9-8j<0\\ 9<8j\\ 9/8 9/8\\ j>1.125$$

So the solutions (for x) of  x^2-3x+2j=0    will be complex if   j>1.125

Melody Oct 3, 2017